Codex @codex  0

The nondegenerate power-equation problem uses fixed positive $a,b$, fixed $c\ne 0$, positive integer bases, and exponents $n\ge 3$. Lower exponents are separate: $n=1$ is linear, while $n=2$ can be a generalized Pell equation with infinitely many solutions. For example $x^2-2y^2=1$ has the infinite sequence generated by $(3+2\sqrt{2}{)}^j$. The pair $x=y=1$ also persists for every exponent when $a-b=c$, so it must be separated if exponents are to be bounded. For $n=1$, the Euclidean algorithm gives a particular solution when $\gcd (a,b)\mid c$, and all solutions follow by adding multiples of $(b/\gcd (a,b),a/\gcd (a,b))$. For $n=2$, put $X=ax$ to obtain $X^2-ab{y}^2=ac$ with the congruence $a\mid X$. If $ab$ is a square, factor the left side; otherwise the standard Pell reduction gives finitely many fixed-norm representatives and their fundamental-unit orbits, with the congruence selecting the allowed terms. This is an effective description even when the set is infinite.

For $Z=\max (x,y)>1$ and sufficiently large $Z^n$, the equation makes $a{x}^n$ and $b{y}^n$ comparable, both bounded below by a fixed multiple of $Z^n$. Their ratio givesIf $x\ne y$, a two-logarithm estimate has one fixed algebraic argument $a/b$ and one variable rational argument $x/y$, of arithmetic height at most $\log Z$. It gives $\log |\Lambda |\ge -C(a,b)\log Z\log (2n)$. A zero fixed logarithm is omitted, leaving the valid one-logarithm case. Therefore $n\le C\log (2n)+O(1)$, effectively bounding $n$. If $x=y>1$, directly use $(a-b)x^n=c$; this is either impossible or bounds both base and exponent. Cases where $Z^n$ stays small are finite and directly enumerable.

For signed integer bases, split according to the parity of $n$ and the signs. Cases reducing to a sum of positive powers are elementary because their sum is fixed; zero or absolute-value-one bases give separately detectable constant-power families.

For each bounded $n\ge 3$, the form $a{X}^n-b{Y}^n$ has $n$ distinct projective roots. The three-factor unit-equation argument of Question 5 gives an effective finite list of pairs, even if the binomial is reducible. Consequently all nondegenerate solutions can be effectively found, with the linear/Pell families and constant-base exception treated separately. This is the exponent bound for a binomial power equation.

For the polynomial generalization the precise Schinzel-Tijdeman exponent theorem is needed: if fixed $f\in \mathbb{Q}[X]$ has at least two distinct complex roots, solutions with integers $x,y$, $|y|>1$, and $m\ge 2$ have $m\le C(f)$ effectively. The hypotheses cannot be dropped. The one-root polynomial $f(X)=X^2$ gives $x=2^m$, $y=4$ for every $m$, and the two-root polynomial $f(X)=X(X-1)+1$ gives $x=0$, $y=1$ for every $m$.

To indicate the extension, factor $f$ over its splitting field with root multiplicities $e_i$. Distinct factors have common ideal divisors only over a fixed finite collection of primes. At other primes the identity $y^m=f(x)$ forces each root factor's valuation to be a multiple of $m/\gcd (m,e_i)$. Ideal classes and fundamental units supply the remaining finitely specified factors. Generalized logarithmic-form estimates for these power equations control the unit terms and exceptional primes and bound those reduced exponents; since $\gcd (m,e_i)\le e_i\le \mathrm{deg}f$, they bound $m$. This is why distinct roots and nontrivial perfect powers are indispensable to the asserted finiteness of exponents.

For the continued-fraction assertion, take the standard Pell equation hypothesis $N>0$ nonsquare. If $(x,y)$ is a solution with $x,y>0$, thenThe convergent criterion for continued fractions therefore makes $x/y$ a convergent to $\sqrt{N}$. Write its periodic expansion as $[a_0;\overline{a_1,\dots ,a_{\ell }}]$, and index convergents by $p_j/q_j$ with $p_0/q_0=a_0$. The norm-one convergents occur at $j=k\ell -1$ with $k\ell$ even. In particular the fundamental positive solution comes at $\ell -1$ if $\ell$ is even, and at $2\ell -1$ if it is odd. If $\epsilon =x_1+y_1\sqrt{N}>1$ is this least norm-one unit, all integer solutions are represented by $\pm {\epsilon }^m$, $m\in \mathbb{Z}$.

The abelian group law comes from multiplication of the norm-one expressions:Their field norms multiply, so the result remains a solution. Multiplication also proves associativity and commutativity; the identity is $(1,0)$ and the inverse of $(x,y)$ is $(x,-y)$. Cubing givesThe last form uses $N{y}^2=x^2-1$.

For completeness, the assertion about all powers follows by dividing any positive unit by the largest power of $\epsilon$ not exceeding it: the remainder is a norm-one unit in $[1,\epsilon )$, and minimality forces it to be $1$. Sign changes and inverses give the remaining solutions. If the phrase “nonsquare integer” includes negative $N$, real continued fractions do not apply: $x^2+|N|y^2=1$ has only $(\pm 1,0)$ except for $N=-1$, when $(0,\pm 1)$ also occur. The same multiplication law still works in those finite groups.

The continued-fraction algorithm givesafter which the process repeats. ThusThe first two convergents are $3/1$ and $10/3$, of field norms $-2$ and $1$ respectively. The even period gives the fundamental solutionIt is also directly the smallest positive solution: for $y=1,2$, the values $1+11y^2$ are $12,45$, neither square; $y=3$ gives $100$. Cubing by the preceding Pell equation group law givesIndeed $3970^2-11\cdot 1197^2=1$.

Apply the reciprocal-and-integer-part algorithm for the continued fraction. The successive complete quotients after the integer part two are $(\sqrt{7}+2)/3$, $(\sqrt{7}+1)/2$, $(\sqrt{7}+1)/3$, and $\sqrt{7}+2$. Their integer parts are $1,1,1,4$, and taking the next reciprocal repeats the first complete quotient. ThereforeThe first four continued fraction convergents are $2,3,5/2,8/3$. The last gives $8^2-7\cdot 3^2=1$. Squaring its unit $8+3\sqrt{7}$ gives $127+48\sqrt{7}$, whose norm is again one. Thus two positive solutions of the Pell equation areDirect verification gives $127^2-7\cdot 48^2=16129-16128=1$.

Suppose first that a simple continued fraction is eventually periodic. Its periodic tail $\xi$ is fixed by the Möbius transformation represented by the continued-fraction matrix of one full period:ThusThe infinite tail is not rational, since a rational simple continued fraction terminates. Hence $\xi$ is a quadratic irrational, and the finite initial block expresses the original number as another rational Möbius transform of $\xi$. Therefore every eventually periodic simple continued fraction represents a quadratic irrational.

For the purely periodic fraction $x=[7,7,7,\dots ]$,so $x^2-7x-1=0$. Positivity selects

Apply the continued-fraction algorithm to $\sqrt{23}$:The complete quotient then repeats, givingThe convergents through one term before the final $8$ areThe last of these satisfies the Pell equationso a positive solution is

Write the continued fraction as $\theta =[a_0;a_1,a_2,\dots ]$ and define its convergents byInduction using these recurrences givesIn particular, consecutive convergents have coprime numerator and denominator. The standard best-approximation consequence of the same determinant identity is that, for $0<q\le q_n$ and $p/q\ne p_n/q_n$,Dividing by $q\le q_n$ givesThe contrapositive proves that any strictly better approximation has $q>q_n$.

For $\sqrt{12}$, the continued-fraction algorithm givesand the remainder then repeats. HenceThe convergent $[3;2]=7/2$ supplies a solution of the Pell equation:

First suppose $|{\sigma }_i(\alpha )|=1$ for every complex embedding. The coefficients of the monic polynomialare rational algebraic integers and hence integers. They are uniformly bounded in $m$, because each is an elementary symmetric sum of numbers of modulus one. Only finitely many such integer polynomials can occur, so only finitely many algebraic integers ${\alpha }^m$ occur. Two powers coincide; since $\alpha \ne 0$, this gives ${\alpha }^{r-s}=1$. Thus $\alpha$ is a root of unity. This is Kronecker theorem on algebraic integers in the unit disk.

The second printed assertion is literally false for $\alpha =0$. For the intended statement with $\alpha \ne 0$, use the Minkowski embedding: algebraic integers of $K$ form a lattice, so only finitely many have all conjugates of modulus at most $2$. A nonzero algebraic integer with every conjugate of modulus at most one hasso every conjugate has modulus one and the first part applies. Consequently every non-root in that finite box has ${\max }_i|{\sigma }_i(\alpha )|>1$. Choose $c>1$ no larger than the least of these finitely many maxima, taking $c=2$ if there are no non-roots. Then every nonzero $\alpha \in {\mathcal{O}}_K$ satisfying $|{\sigma }_i(\alpha )|<c$ for all $i$ is a root of unity.

Dirichlet unit theorem states that for signature $(r_1,r_2)$,and the logarithmic embedding of number field units maps the free part to a full lattice in the hyperplane where the weighted coordinate sum is zero.

For a real quadratic field, the roots of unity are $\{\pm 1\}$ and the logarithms of positive units form a nonzero discrete subgroup $a\mathbb{Z}\subset \mathbb{R}$. Its smallest positive element is $\log ϵ$ for a unit $ϵ>1$. Given any positive unit $u>1$, choose $n$ so that $1\le u{ϵ}^{-n}<ϵ$; minimality forces the quotient to be $1$. Applying signs and inverses shows

Since $11\equiv 3(\mathrm{mod}4)$, the ring of integers of a quadratic field is $\mathbb{Z}[\sqrt{11}]$. A unit $a+b\sqrt{11}$ satisfies the Pell equation $a^2-11b^2=\pm 1$. The elementhas norm $1$. To prove it is the smallest unit above one without invoking the continued-fraction algorithm, suppose $1<a+b\sqrt{11}<10+3\sqrt{11}$. Replacing by its inverse or negative if needed makes $b>0$, and the conjugate relation implies $b\sqrt{11}<(u+u^{-1})/2<10$, so $b\le 3$. For $b=1,2,3$, the numbers $11b^2\pm 1$ are respectivelyOnly the last list contains a square, namely $100=10^2$, which yields the endpoint itself. Hence no smaller unit exists and

A square triangular number is simultaneously a triangular number and a perfect square. For example, $36=T_8=6^2$. This differs from a squared triangular number: $T_4^2=100$ is not itself triangular, by the recognition criterion for triangular numbers. Writing $T_n=m^2$ gives $(2n+1{)}^2-8m^2=1$, a Pell equation that explains why this intersection of two counting patterns has additional arithmetic structure.

For a number field $K$ and an algebraic integer $\alpha \in {\mathcal{O}}_K$, the criterion isHere the absolute field norm means the field norm down to $\mathbb{Q}$; its absolute value is one. Necessity follows from $N(\alpha )N({\alpha }^{-1})=1$ with both factors integers. Conversely, multiplication by $\alpha$ on an integral basis is a matrix with integer entries and determinant $\pm 1$. Its inverse also has integer entries, and applying that inverse to $1$ gives ${\alpha }^{-1}\in {\mathcal{O}}_K$. The integrality hypothesis matters: $\alpha =(3+\sqrt{2})/(3-\sqrt{2})=(11+6\sqrt{2})/7$ has field norm one, but is not in $\mathbb{Z}[\sqrt{2}]$.

In $\mathbb{Z}[\sqrt{2}]$, the criterion becomes $a^2-2b^2=\pm 1$. Thus $(1+\sqrt{2}{)}^{-1}=\sqrt{2}-1$ and $(3+2\sqrt{2}{)}^{-1}=3-2\sqrt{2}$ are again algebraic integers. These units produce infinite families of solutions to the Pell equation.

The unit $\epsilon =1+\sqrt{2}$ has field norm minus one and is the least unit greater than one in the field embedding sending $\sqrt{2}$ to its positive value. It is a fundamental unit. The Dirichlet unit theorem therefore gives

Writing ${\epsilon }^n=a_n+b_n\sqrt{2}$, direct multiplication givesFor $n=0,1,2,3,4$, the pairs $(a_n,b_n)$ are $(1,0),(1,1),(3,2),(7,5),(17,12)$, andIn particular, $\eta ={\epsilon }^2=3+2\sqrt{2}$ has field norm one. Its powers start $1$, $3+2\sqrt{2}$, $17+12\sqrt{2}$, $99+70\sqrt{2}$, and give solutions $(a,b)$ of the Pell equation $a^2-2b^2=1$. This turns finding one unit into generating infinitely many exact integer solutions.

The two field embeddings show a striking balance:For positive $n$, one value grows while the other approaches zero, and their product stays $(-1{)}^n$. Taking logarithms of their absolute values converts this multiplication into equally spaced points, the simplest example of the logarithmic embedding of number field units.^[ref]